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anonymous
 5 years ago
Gasoline is pouring into a cylindrical tank of radius 3 feet. When the depth of the gasoline is 6 feet, the depth is increasing at 0.2 ft/sec. How fast is the volume of gasoline changing at that instant?
Round your answer to three decimal places.
anonymous
 5 years ago
Gasoline is pouring into a cylindrical tank of radius 3 feet. When the depth of the gasoline is 6 feet, the depth is increasing at 0.2 ft/sec. How fast is the volume of gasoline changing at that instant? Round your answer to three decimal places.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, you can find the volume first since you have h and r , so : \[V = \pi r^2 h\] \[= ( 3)^2 (6) \pi\] = 169.65 ft^3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sstarica, isn't this the same thing we were tackling

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now find V' which is :\[V' = 2\pi rr'h'\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not really @iam , somehow similar :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now we have: h' = 0.2 ft/sec we need to find r'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes :) but we need to find r' now

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[V' = 2 \pi rr' h'\] we have h' , but we need to find r' to calculate V' = rate of volume increase

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you need to find a relationship between r and h so r/h = r'/h'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats where im having problem for all the problems

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0cross mulitply and you'll get: r' = rh'/r

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry! wait you'll get: r' = rh'/h :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you with me soo far?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now plug : r =3 ft h = 6ft h' = 0.2ft/sec r' = (3)(0.2)/6 = 0.1 ft/sec

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we have found r' now, let's go back to V'\[V' = 2 \pi rr'h'\] so, V' = 2(3)(0.1)(0.2) pi = 0.377 ft^3/sec

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Correct me if I'm wrong please

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0OMG this makes so much sense now

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but the website marked it wrong :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I guess there's a mistake, but that's the idea, maybe in calculation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did you get the idea?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where did u get \[V \prime = 2\pi r r \prime h \prime \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you derive with respect to r and h :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you want to know how is the volume increasing when both the radius and height is changing, that's the story of the related rated problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oooooh...im sorry can u show me one more problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmmm, let's say we have a circle of radius = 2 m, and the radius is decreasing at 0.5 m/sec. How much will the circle's area increase ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'll give you advise, whenever you see ( m/sec) or anything with respect to time it's something' which in this case is the r' = 0.5m/sec

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so what is the given?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we have :  R = 2 m  R' = 0.5 m/sec and we know the formula of the Area of the circle which is: A = pi r^2, right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now , you want to know how much is the area increasing with respect to R, since R is increasing :) so you find the derivative of it which will be : A' = 2 pi rr'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so when exactly do i start pluging in numbers?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right now, after you find the derived formula, start pluggingthe numbers :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay ...these are pretty simple compared to the ones i have on hw :'(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Like this problem, they want to know how much is the area increasing , we got the formula so the answer is: A' = 2(2)(0.5) pi = 2 pi = 2.68 m^2/sec :) So the area is increasing at 2.68 m^2/sec

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Lol, when you understand the small idea, you'll be able to figure out the bigger idea :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did you understand my problem?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes but how would i approach towards a hemisphere problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0think about it, write down the given on one side, and the RTF(required to find) on the other side, then SKETCH! To test your understanding , you've got to sketch what they give you to be fully understand what they want :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01) write the given and what you need to find 2) sketch 3) write down the general formula that you're going to use :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0good luck ! You're a smart student, you can figure it out :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you got the small idea, you'll be able to solve any bigger ideas now . They are not even big, they are small ideas that got stretched ^_^, don't let the problem boss u, YOU ARE THE BOSS!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0np :) Give it a try once more. ^_^
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